Kinematics - Acceleration in Two Dimensions. Stuck.

In summary, the conversation is about someone having trouble with a physics question involving a hockey puck rebounding from a board and determining its average acceleration. They have tried using the cosine law and vector components, but are unable to get the correct answer. They also mention a lawn mower example from the same textbook that they are also struggling with. They are seeking help to understand the concepts and solve the problem correctly.
  • #1
mattstjean
14
0
Hi, I am also having trouble with the hockey puck question.

A hockey puck rebounds from a board as shown in my diagram. The puck is in contact with the board for 2.5 ms. Determine avg acceleration of the puck over the interval.

Vi = 26 m/s Vf = 21 m/s

I tried the cosine law but I keep getting 44 m/s and not 18. I don't understand how you guys got 18 m/s. I've plugged it in at least 100 times as
[itex]
v_t = \sqrt{v_1^2 + v_2^2 - 2(v_1)(v_2)cos136}
= \sqrt{26^2 + 21^2 - 2(26)(21)cos136}
=44
= [/itex]

Because that wasn't working I then tried Vector Components and I can't get that to work either. I did:
[itex]
V_x = V_B sin \theta + (-V_A cos \beta )
= 21 sin(22) - 26 cos(22)
=-16
[/itex]

and

[itex]
V_y = V_B cos \theta + (-V_A sin \beta )
= 21 (cos22) + 26(sin22)
= 29
[/itex]

I then tried to figure out
[itex]
\Delta V ^2= \Delta V_x ^2 + \Delta V_y^2
= sqrt{16^2 + 29^2}
= 33
[/itex]

Using that I tried to get the average acceleration by:

[itex]
A_av = \Delta V / \Delta T

A_av = 33 / 2.5x10^-3
A_av = 13.2x10^3
[/itex]
and to find the angle I tried to do :

[itex]
\phi = tan^-1 = 16/29
\phi = 29degrees
[/itex]

However, the answer in my book says that the average acceleration is [itex] 7.3x10^3 [7.5degrees North of West][/itex]
Any help would be amazingly appreciated. Thanks.
 

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  • #2
mattstjean said:
Because that wasn't working I then tried Vector Components and I can't get that to work either. I did:
[itex]
V_x = V_B sin \theta + (-V_A cos \beta )
= 21 sin(22) - 26 cos(22)
=-16
[/itex]
You have a mix of sine and cosine. Only one is correct.
and

[itex]
V_y = V_B cos \theta + (-V_A sin \beta )
= 21 (cos22) + 26(sin22)
= 29
[/itex]
Again, a mix of sine and cosine.

Redo this.
 
  • #3
Doc Al said:
You have a mix of sine and cosine. Only one is correct.

Again, a mix of sine and cosine.

Redo this.

I don't know how to redo it. In my textbook it used them both together in the y and x component vector subtraction. I took the equations right out of my text, Nelson Physics 12.
 
  • #4
mattstjean said:
I don't know how to redo it. In my textbook it used them both together in the y and x component vector subtraction. I took the equations right out of my text, Nelson Physics 12.
I'm not sure what equations you are talking about.

Do this: What's the x-component of Vi? The x-component of Vf?
 
  • #5
I have the same book and I am stuck on the example on right before the quesion box you asked about. can you please explain the lawn mower example in pg 28. I AM VERY CONFUSED
 

Related to Kinematics - Acceleration in Two Dimensions. Stuck.

1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion.

2. What is acceleration in two dimensions?

Acceleration in two dimensions refers to the change in velocity of an object in two different directions, typically represented by the x and y axes.

3. How is acceleration in two dimensions calculated?

Acceleration in two dimensions is calculated by dividing the change in velocity in each direction by the change in time.

4. What are some examples of two-dimensional acceleration?

Some examples of two-dimensional acceleration include a ball rolling down a ramp, a plane taking off or landing, and a car making a turn.

5. How does acceleration in two dimensions differ from one-dimensional acceleration?

In one-dimensional acceleration, the object is only moving in one direction, so the acceleration is only measured in that direction. In two-dimensional acceleration, the object is moving in two directions, so the acceleration must be calculated for each direction separately.

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